How do you find the critical numbers of # y= x^2*(1 + 3 ln x)#?

1 Answer
Apr 2, 2018

Answer:

#x=e^(-5/6)#

Explanation:

Differentiate, set equal to zero, and solve for #x,# in addition to being aware of specific values for which the derivative doesn't exist.

#y'=x^2*d/dx(1+3lnx)+(1+3lnx)*d/dxx^2#

#y'=x^2(3/x)+2x(1+3lnx)#

#y'=3x+2x+6xlnx#

#y'=5x+6xlnx#

The derivative is logarithmic like the original function. They both share the domain #(0, oo),# so there aren't any particular values for which the derivative doesn't exist while the function does.

#5x+6xlnx=0#

#x(5+6lnx)=0#

#x=0# looks like it would be a solution, but since it's not in the domain of the logarithm, it is not valid.

#5+6lnx=0#

#6lnx=-5#

#lnx=-5/6#

#e^lnx=e^(-5/6)#

From the fact that #e^lnx=x:#

#x=e^(-5/6)#

Is the only critical number.